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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Solve system of linear equations, using matrix method:\[\] \[2x-y=-2\] \[3x+4y = 3\]

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Toolbox:
  • A matrix is said to be singular if |A|= 0.
  • A matrix is said to be invertible if |A|$\neq 0$.
  • If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
  • Using this we can solve the system of equation which has unique solution.
This can be written in the form AX=B.
 
$\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\3\end{bmatrix}$
 
Where $A=\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}\;X=\begin{bmatrix}x\\y\end{bmatrix}\;B=\begin{bmatrix}-2\\3\end{bmatrix}$
 
Let us evaluate $|A|=2\times 4-(-1)\times 3$
 
$\qquad\qquad\qquad=8+3=11.$
 
$|A|\neq 0.$
 
Hence it is non singular.
 
Therefore inverse exists.
 
Now let us find out adj A.
 
This can be obtained by interchanging the element of $a_{11}$ and $a_{22}$ and interchanging th symbols of elements of $a_{21}$ and $a_{12}$
 
Therefore $adj A=\begin{bmatrix}4 & 1\\-3 & 2\end{bmatrix}$ and |A|=11.
 
$A^{-1}=\frac{1}{11}\begin{bmatrix}4 & 1\\-3 & 2\end{bmatrix}$
 
Matrix multiplication can be obtained by multiplying the rows of A with the column of B.
 
Now AX=B$\Rightarrow X=A^{-1}B.$
 
$x=\frac{1}{11}\begin{bmatrix}4 & 1\\-3 & 2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}$
 
$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-8+3\\6+6\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}$
 
$\Rightarrow \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}-5/11\\12/11\end{bmatrix}$
 
Hence x=-5/11 and y=12/11.

 

answered Feb 27, 2013 by sreemathi.v
edited Feb 27, 2013 by vijayalakshmi_ramakrishnans
 

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